- Trig angle addition identities
- Using the cosine angle addition identity
- Using the cosine double-angle identity
- Using the trig angle addition identities
- Proof of the sine angle addition identity
- Proof of the cosine angle addition identity
- Proof of the tangent angle sum and difference identities
Sal proves the identity sin(x+y) = sin(x)*cos(y) + cos(x)*sin(y). Created by Sal Khan.
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- I understand how this video proves the angle addition for sine, but not where this formula comes from to begin with, I feel like somewhere I missed a step. It seems like a very complex proof for such a simple concept, why can't we just add sine a + sine b directly? What is it about trig functions that makes angle additions so complicated? I felt like I was grasping all the trig identities/unit circle definitions, etc up to this point but just crashed & burned here... The video is very clear but it seems like there should be some sort of introductory video to the concept of adding angles & why we can't do it more easily... Maybe it's just me?(62 votes)
- The proof is where the formula comes from. We can't do sin(a + b) = sin(a) + sin(b) because sine does not distribute. It's similar to x^2: (a + b)^2 isn't a^2 + b^2, it's a^2 + 2ab + b^2. The same thing applies to sin(a + b): sin(a + b) = sin(a)cos(b) + cos(a)sin(b).(101 votes)
- It seems that this (very nice) proof only covers the case where x, y, and (x+y) are all acute angles. How would one generalize this to cover all x and y?(43 votes)
- This is a good question. Here's a proof I just came up with that the angle addition formula for sin() applies to angles in the second quadrant:
Given: pi/2 < a < pi and pi/2 < b < pi // a and b are obtuse angles less than 180°.
Define: c = a - pi/2 and d = b - pi/2 // c and d are acute angles.
Theorem: sin(c + d) = sin(c)*cos(d) + cos(c)*sin(d) // angle addition formula for sin().
Substitute: sin((a - pi/2) + (b - pi/2)) = sin(a - pi/2)*cos(b - pi/2) + cos(a - pi/2)*sin(b - pi/2)
Simplify: sin((a - pi/2) + (b - pi/2)) = sin(a + b - pi)
sin(a + b - pi) = -sin(a + b) // from unit circle
sin(a - pi/2) = -cos(a) and sin(b - pi/2) = -cos(b) // from unit circle
cos(a - pi/2) = sin(a) and cos(b - pi/2) = sin(b) // from unit circle
Substitute: -sin(a + b) = -cos(a)*sin(b) + sin(a)*(-cos(b))
Simplify and rearrange: sin(a + b) = sin(a)*cos(b) + cos(a)*sin(b) // Done
You can also use this method to generalize to angles greater than 180°, and it also works for the cos() addition formula.(26 votes)
- I understood all the parts of the proof, but is it okay to generalize this after proving it with only a right triangle that has a side of unit measure? Isn't there a way to prove it holds true for all angles, without taking the measure of a side as 1?(11 votes)
- The trigonometric functions are all based on the ratios of the sides of right triangles and for a similar triangle (which is the case for triangles with the same angles) this ratio will remain constant as you scale up or down the length of the hypotenuse.(11 votes)
- I tried to post this as a "mistake", but when I hit send, it just hung with the little status indicator in constant motion:
Current convention is that the end point names with a horizontal bar over them names the segment that is the geometric figure defined as the two points plus all points between them on the line that passes through the two points. The end point names with no horizontal bar over them names the segment's length, which is a number. Horizontal line names something spatial; no horizontal line names something quantitative. In the video, horizontal bars are used counter to this convention to name segment lengths.(13 votes)
- at6:43why is that angle labeled 90 - y?(4 votes)
- We don't know what y is, but we know that those two angles add up to 90. That means the other one has to be y less than 90.
Think about it:
y + (that angle) = 90
(that angle) = 90 - y(8 votes)
- Very clear and detailed proof, but where does this diagram even come from? I feel like I need to know where there diagram is derived from as well.(6 votes)
- Hello Omar,
If you have a protractor I encourage you to construct the triangles to show the sum of SIN(30 degrees) plus SIN(45 degrees).
Tips: Let the length of the upper triangle's hypotenuse equal one. Observe that this is the only length that equals one. Take your time to do the procedure even if you need start over multiple times like I did. It's worth the effort and you will see how the triangles came to be. You will also have a nice drawing you can hang on you wall because you will "own" the identity.
Please leave a comment and let us know if you were successful.
- sin(∠ABC−60∘) , im confused whether I should be subtracting or adding the two because I'm subtracting -60∘ instead of adding 60∘ : sin(theta)*cos(60) + cos(theta)*sin(60)
or: sin(theta)*cos(60) - cos(theta)*sin(60)
Do i always add sinacosb and cosasinb regardless of positive or negative value of 'b' in sin(a+b)??
Also what about for cos(a+b)? ( Where the value of b is negative)
Sorry if this doesn't make any sense...(2 votes)
- Addition Formulas:
Cos(a+b)=cos(a)cos(b) - sin(a)sin(b).
Sin(a-b)=sin(a)cos(b) - cos(a)sin(b).
- How to prove sin0/cos0+cos0/sin0=sec0 cos ec0?
I mean i'm putting 0 because i couldnt type theater sign..(1 vote)
- Ok, first the variable is called theta (θ).
As for your problem:
sin(θ)/cos(θ) + cos(θ)/sin(θ) = sec(θ)•csc(θ)
sin(θ)/cos(θ) + cos(θ)/sin(θ) = 1/cos(θ)•1/sin(θ)
sin^2(θ)/cos(θ) + cos(θ) = 1/cos(θ)•sin(θ)/sin(θ)
sin^2(θ) + cos^2(θ) = cos(θ)/cos(θ)•sin(θ)/sin(θ)
sin^2(θ) + cos^2(θ) = 1•1
sin^2(θ) + cos^2(θ) = 1 QED(10 votes)
- Not sure where to ask, but where can I find video/article(s) about these formulas?
sin(x) + sin(y) = 2sin((x+y)/2)*cos((x-y))/2)
sin(x) - sin(y) = 2cos((x+y)/2)*sin((x-y))/2)
cos(x) + cos(y) = 2cos((x+y)/2)*cos((x-y))/2)
cos(x) - cos(y) = -2sin((x+y)/2)*sin((x-y))/2)(4 votes)
- We can derive this formulas by using basic addition identities of sine and cosine .I tried to search them, but i could not find them here.But they are on youtube
Make sure you watch all four videos from 35 to 38(1 vote)
- Say we are using a unit circle. Say A is the center of that unit circle. And we want the proof related that senario. In that case AB is as above 1 but AC would also be 1.... So the above proof would need to be thought of in a different way. Is there any proof of the sine addition identity using the unit circle?(3 votes)
Voiceover: What I hope to do in this video is prove the angle addition formula for sine, or in particular prove that the sine of x plus y is equal to the sine of x times the cosine of -- I forgot my x. Sine of x times the cosine of y plus cosine of x times the sine of y. The way I'm going to do it is with this diagram right over here. You can view it as, it has this red right triangle. It has this right triangle that has a hypotenuse of one. You could say this triangle ADC. It has it stacked on top of, its base is the hypotenuse of triangle ACD, which I could, I'm going to outline it in blue since I already labelled the measure of this angle as being y. AC which is the base of triangle ADC is the hypotenuse of triangle ABC. They're stacked on top of each other like that, just like that. The way I'm going to think about it is first if you just look at this what is sine of x plus y going to be? Well x plus y is this entire angle right over here. If you look at this right triangle, right triangle ADF, we know that the sine of an angle is the opposite side over the hypotenuse. Well the hypotenuse here is one so the sine of this is the opposite over one, or the sine of this angle, the sine of x plus y is equal to the length of this opposite side. So sine of x plus y is going to be equal to the length of segment DF. What I'm going to try to do is, okay, length of segment DF is essentially what we're looking for but we can decompose the length of segment DF into two segments. We can decompose it into length of segment DE and the length of segment EF. We can say that DF, which is the same thing as sine of x plus y, the length of segment DF is the same thing, is equal to the length of segment DE plus the length of segment EF. EF is of course the same thing as the length of segment CB. ECBF, this right over here is a rectangle so EF is the same thing as CB. So this thing is going to be equal to DE right over here, length of segment DE plus the length of segment CB. Once again the way I'm going to address this, the sine of x plus y which is the length of DF and DF can be decomposed as the lengths of DE and CB. Now with that as a hint I encourage you to figure out what the length of segment DE is in terms of x's and y's and sines and cosines, and also figure out what the length of segment CB is in terms of x's and y's and sines and cosines. Try to figure out as much as you can about this and these two might fall out of that. I'm assuming you've given a go at it so now that we know that sine of x plus y can be expressed this way, let's see if we can figure these things out. I'm going to try to address it by figuring out as many lengths and angles here as I can. Let's go to this top red triangle right over here. Its hypotenuse has length one so what's going to be the length of segment DC? That is the opposite side of our angle x so we know sine of x is equal to DC over one, or DC over one is just DC. This length right over here is sine of x. Segment AC, same exact logic. Cosine of x is the length of AC over one which is just the length of AC. This length right over here, segment AC its length is cosine of x. That's kind of interesting. Now let's see what we can figure out about this triangle, triangle ACB right over here. How could we figure out CB? Well we know that sine of y, let me write this here. Sine of y is equal to what? It's equal to the length of segment CB over the hypotenuse. The hypotenuse here is the cosine of x, and I think you might see where all of this is leading. At any point if you get excited pause the video and try to finish the proof on your own. The length of segment CB if we just multiply both sides by cosine of x, the length of segment CB is equal to cosine of x times sine of y. Which is neat because we just showed that this thing right over here is equal to this thing right over here. To complete our proof we just need to prove that this thing is equal to this thing right over there. If that's equal to that and that's equal to that well we already know that the sum of these is equal to the length of DF which is sine of x plus y. Let's see if we can figure out, if we can express DE somehow. What angle would be useful? If somehow we could figure out this angle up here or maybe this angle, well let's see. If we could figure out this angle then DE we could express in terms of this angle and sine of x. Let's see if we can figure out that angle. We know this is angle y over here and we also know that this is a right angle. EC is parallel to AB so you could view AC as a transversal. If this is angle y right over here then we know this is also angle y. These are once again, notice. If AC is a transversal here and EC and AB are parallel then if this is y then that is y. If that's y then this is 90 minus y. If this is 90 degrees and this is 90 minus y then these two angles combined add up to 180 minus y, and if all three of these add up to 180 then this thing up here must be equal to y. Validate that. y plus 90 minus y plus 90 is 180 degrees, and that is useful for us because now we can express segment DE in terms of y and sine of x. What is DE to y? It's the adjacent angle, so we can think of cosine. We know that the cosine of angle y, if we look at triangle DEC right over here, we know that the cosine of y is equal to segment DE over its hypotenuse, over sine of x. You should be getting excited right about now because we've just shown, if we multiply both sides by sine of x, we've just shown that DE is equal to sine of x times cosine of y. We've now shown that this is equal to this. We already showed that CB is equal to that, so the sum of DE and CB which is the same thing as the sum of DE and EF is the sine of x plus y which is that over there. We are done, we have proven the angle addition formula for sine.